Transcript PowerPoint 簡報 - National Chung Cheng University

Chap 6 Residues and Poles
Cauchy-Goursat Theorem:
c f dz  0
if f analytic.
What if f is not analytic at finite number of points interior
to C
Residues.
53. Residues
z0 is called a singular point of a function f if f fails to be analytic at
z0 but is analytic at some point in every neighborhood of z0.
A singular point z0 is said to be isolated if, in addition, there is a
deleted neighborhood 0  z  z0  ε of z0 throughout which f
is analytic.
tch-prob
1
Ex1.
z 1
z 2(z2 1 )
has isolated singnlar points z  0,  i
Ex2. The origin is a singular point of Log z, but is not isolated
Ex3.
1
sin( z)
singular points z  0 and z  1n n  1,  2, ....
not isolated
isolated
When z0 is an isolated singular point of a function f, there is a R2
such that f is analytic in 0  z  z0  R2
tch-prob
2
Consequently, f(z) is represented by a Laurent series

bn
b
b2
f ( z)   an ( z  z0 )n  z 1z 

.....

 .......
(1)
2
n
(
z

z
)
(
z

z
)
0
0
0
n0
0  z  z0  R2
where bn  1 c f ( z)dz
(n 1, 2, ... )
2 i (z  z )n1
0
and C is positively oriented simple closed contour
around z0 and lying in 0  z  z0  R2
When n=1, 2πi b1  c f (z) dz
(2)
1
The complex number b1, which is the coefficient of z  z in
0
expansion (1) , is called the residue of f at the isolated singular
point z0.
A powerful tool for evaluating certain
Re s f ( z)
z  z0
integrals.
tch-prob
3
c
Ex4.
dz
z( z  2)4
C : z- 2 1
1
has singular points at z  0, z  2
z( z  2)4
it has Laurent series representation
in 0  z - 2  2
dz  2πi Re s
1
based on (2)
c
z  2 z( z  2)4
z( z  2)4
but
1  1 .
1
z( z - 2)4 ( z  2)4 2  ( z  2)
1
 1 .
2( z - 2)4 1 ( z  2)
2
 (-1)n
 
(z - 2)n-4
n0 2n1
b1   1
16
c
0
2
湊出z-2在分母
0 z -2  2
dz  2 i   1   π i
 16  8
z( z - 2)2


tch-prob
4
Ex5. show c exp ( 12 ) dz  0
z
where
C : z 1
1 is analytic everywhere except at the origin
z2
2 3
e z 1 z  z  z  ......
z 
1! 2! 3!
1
e z2 1 1  1 1  1 1  ..... 0  z  
1! z 2 2! z 4 3! z6
b1  0
c exp ( 1 ) dz  0
z2
f analytic on and within C  c f dz  0
The reverse is not necessarily true.
tch-prob
5
54. Residue Theorems
Thm1. Let C be a positively oriented simple closed contour. If f
is analytic inside and on C except for a finite number of (isolated)
singular points zk inside C, then
n
f ( z)
c f ( z) dz  2πi  Res
z  zk
k 1
pf:
but
n
f
(
z
)
dz

 c f ( z) dz  0
c
k 1 k
c f ( z) dz  2π i Res f ( z)
k
z  zk
tch-prob
Cauchy’s residue theorem
C
Z3
Z1
Z2
6
Ex1.
Evaluate c 5z - 2 dz
where C : z  2
z(z -1 )
Two singularities z  0, z 1
a. When 0  z 1
5z - 2  5z  2 -1  (5  2) (-1- z - z 2.....)
z
z
1- z
z( z -1)
b1  B1  Re s f ( z)  2
z 0
b. when 0  z-1 1
5z- 2  5( z 1)  3 .
1
z 1
z( z-1)
1 ( z 1)
 (5  3 ) [1- ( z-1)  ( z-1)2.......]
z 1
b1  B2  3
c 5z  2 dz  2π i (2  3) 10π i
z( z 1)
tch-prob
7
Thm2: If a function f is analytic everywhere in the finite plane except
for a finite number of singular points interior to a positively
oriented simple closed contour C, then
C0
1
1
s [ 2 f ( )]
c f (z) dz  2πi Re
z 0
z
z
Pf:
From Laurent Theorem

f ( z)   cn z n
(R1  z  ) (3)
n
where cn  1 c f ( z) dz
2πi 0 zn1
c0 f ( z) dz  2πi c1 (c1 is not the residue of f at z =0)
C
R1 R0
Replace z by 1 in (3),
z

 c
cn
1 1
1
n2
f
(
)


(0

z

)


2
n2
n
z
z n
z
R1
n z
now c1 is the residue
1
1
c1  Re s [ 2 f ( )]
1 f (1) at z  0
z 0
z
z
of
tch-prob
z2 z
8
Ex2.
f ( z)  5z  2
z  ( z 1)
1 f (1 )  5  2 z  5  2 z . 1
z 1 z
z 2 z z(1 z)
 (5z  2)(1 z  z 2  ....)
 5z  3  3z  ......
Re s 1 f (1)  5
z
z 0 z 2
c f ( z)dz 10πi
tch-prob
(0  z 1)
9
55. Three Types of Isolated Singular points
If f has an isolated singular point z0, then f(z) can be represented by
a Laurent series

bn
b1
b2
n
f ( z)   an( z  z0 )  z  z 
.....
....
2
n
( z  z0 )
0 ( z  z0 )
n0
in a punctured disk 0  z  z0  R2
bn
b1
b2


.....

....
n
z  z0 ( z  z )2
( z  z0 )
0
is called the principal part of f at z0.
The portion
tch-prob
10
(i) Type 1.
bm  0 and bm1  bm 2  ....... 0

bm
b
f ( z)   an ( z  z0 )n  1  ..............
m
(
z

z
)
(
z

z
)
0
n0
0
0  z  z0  R2
The isolated singular point z0 is called a pole of order m.
m  1,  simple pole
Ex1.
z2  2z  3  z  3  2  ( z  2)  3
z 2
z 2
z 2
(0  z- 2 )
Simple pole m 1 at z0  2, b1  3.
tch-prob
11
Ex2.
sinh z  1 ( z  z3  z5  ....)
3! 5!
z4
z4
3
1
1
1
z
z
 3  z    ...........0  z  
5! 7!
z 3!
has pole of order m  3 at z0  0, b1  1
6
(ii) Type 2
bn=0, n=1, 2, 3,……

f ( z)   an ( z  z0 )n  a0  a1 ( z  z0 )  a2 ( z  z0 )2  ......
n0
0  z  z0  R2
z0
is known as a removable singular point.
* Residue at a removable singular point is always zero.
tch-prob
12
* If we redefine f at z0 so that f(z0)=a0
define
Above expansion becomes valid throughout the entire disk
z  z0  R2
* Since a power series always represents an analytic function
Interior to its circle of convergence (sec. 49), f is analytic
at z0 when it is assigned the value a0 there. The singularity
at z0 is therefore removed.
Ex3.
2 4 6
f ( z)  1 cos z  1 [1 (1 z  z  z  ....)]
2! 4! 6!
z2
z2
2 4
 1  z  z  .......
(0  z )
2! 4! 6!
when the value f (0)  1 is assigned, f become entire,
2
the point z0  0 is a removable singular point.
* another example f ( z)  sinz z .
tch-prob
13
(iii) Type 3:
Infinite number of bn is nonzero.
z0 is said to be an essential singular point of f.
In each neighborhood of an essential singular point, a
function assumes every finite value, with one possible
exception, an infinite number of times. ~ Picard’s theorem.
tch-prob
14
Ex4.

exp(1z )   1 1n 1 1 1z  1 12  ...... 0  z  
1! 2! z
n0 n! z
has an essential singular point at z0  0
where the residue b1  1
* Note that exp z  -1 when z  (2n 1)π i
(n  0, 1,  2, ...)
1
i
 exp (1z )  1 when z 
(2n 1)πi (2n 1)π
(n  0, 1,  2, ...)
an infinite number of these points clearly lie in any given
neighborhood of the origin.
* Since exp(1z )  0 for any value of z, zero is the exceptional
value in Picard's theorem.
tch-prob
15
* exp z 1 when z  2nπ i
(n  0, 1,  2, ...)
 exp (1z ) 1 when z  1  - i
2nπi 2nπ
(n  0, 1,  2, ...)
* exp z  i when z  (2n 1/2)π i
(n  0, 1,  2, ...)
1
i
 exp (1z )  i when z 
(2n 1/2)πi (2n 1/2)π
(n  0, 1,  2, ...)
an infinite number of these points clearly lie in any given
neighborhood of the origin.
tch-prob
16
56. Residues at Poles
identify poles and find its corresponding residues.
Thm. An isolated singular point z0 of a function f is a pole
of order m iff f(z) can be written as
f ( z)   ( z) m
( z  z0 )
where  ( z) is analytic and nonzero at z0 .
Moreover, Res f ( z)   ( z0 ) if m 1
z  z0
and
Res f ( z) 
z  z0
tch-prob
 (m-1) ( z0 )
(m 1)!
if m  2
17
Pf: “<=“
Suppose f ( z)   ( z) m .
( z - z0 )
Since  ( z) is analytic at z0, it has a Taylor series representation
 ( z)   ( z0 )   '( z0 ) ( z  z0 )   ''( z0 ) ( z  z0 )2  .......
1!
2!
(m 1)
  ( n) ( z )

( z0 )
0

( z  z0 )m1  
( z  z0 ) n
n!
(m 1)!
nm
 ( z0 )   '( z0 ) /1!   ''( z0 ) / 2!  .......
f ( z) 
( z  z0 ) m ( z  z0 ) m1 ( z  z0 ) m 2
(m 1)

( z0 ) /(m 1)!




 ( n ) ( z0 )
( z  z0 ) n  m
n!
( z  z0 )
nm
Since  ( z0 )  0, z0 is a pole of order m of f ( z) and
z  z0  
0 z  z0  
(m 1)

( z0 )
Res f ( z) 
.
z  z0
(m 1)!
tch-prob
18
“=>”
If z0 is a pole of order m of f , or f ( z) has a Laurent series representation

b
b2
bm
f ( z)   an( z  z0 )n  z 1z 

.....

(bm  0)
2
m
(
z

z
)
(
z

z
)
0
n0
0
0
in a punctured disk 0  z  z0  R2
The function defined by
( z  z0 ) m f ( z ) when z  z0
 ( z)  
bm
when z  z0

has the power series representation
 (z)  bm  bm1(z  z0 )   b2 (z  z
(z  z0
)m2  b1
0
)m1 

 an ( z  z0 )
m n
n 0
throughout z  z0  R2.
Consequently,  ( z) is analytic in that disk (sec.49)
and, in particular at z0.
Also  (z0 )  bm  0.
tch-prob
19
Ex1.
f (z)  z 1 has an isolated singular point at z  3 i
z2  9
f ( z)   ( z)
z 3 i
where  ( z)  z 1
z 3 i
 ( z) is analytic at z  3 i
 (3 i)  3 i 1  0 a simple pole
6i
Re s  3  i
6
z 3i
another simple pole
z  -3 i
residue 3  i
6
tch-prob
20
Ex3.
f ( z)  sinh z
z4
To find residue at z0  0,
can not write f ( z)   ( z) ,  ( z)  sinh z
z4
since  ( z0 )  0
Need to write out the Laurent series for f(z) as in Ex 2.
Sec. 55.
sinh z  1  1 1  1 z  ..........
z4
z3 3! z 5!
z0  0 is a pole of the third order, its residue  1
6
tch-prob
21
Ex4.
Since z(e z 1) is entire and its zeros are
z  2n i (n  0, 1,  2, ..... )
z  0 is an isolated singular point of
f ( z)  z1
z(e 1)
2 3
e z 1 z  z  z  ....
z 
1! 2! 3!
2
z(e z 1)  z 2(1 z  z  .....) z  
2! 3!
1
Thus f ( z)   ( z)
 ( z) 
2
z2
1 z  z  .......
2! 3!
Since  ( z) is analytic at z  0, and  (0) 1  0
z  0 is a pole of the second order
( 1  2 z  .....)
2! 3!
b1  ' (0) 
2
(1 z  z  .....)2
z 0
2! 3!
tch-prob
-1
2
22
57. Zeros and Poles of order m
Consider a function f that is analytic at a point z0.
(From Sec. 40). f (n)( z) (n 1, 2, ....) exist at z
0
If f ( z0 )  0 ,
f '( z0 )  0
:
f (m-1) ( z )  0
0
f (m) ( z0 )  0
Then f is said to have a zero of order m at z0.
Lemma:
f ( z)  ( z  z0 )m g ( z)
analytic and non-zero at z0.
tch-prob
23
Ex1.
f ( z)  z(e z 1)
2
 z 2(1 z  z  ......)
2! 3!
has a zero of order m  2 at z0  0
(e z 1) / z when z 0
g ( z)  
when z 0
 1
is analytic at z  0.
Thm. Functions p and q are analytic at z0, and p(z0)  0.
If q has a zero of order m at z0, then
p( z) has a pole of order m there.
q( z)
q( z)  ( z  z )m g ( z)
0
analytic and non zero
p( z)  p( z) /g ( z)
q( z) ( z  z0 )m
tch-prob
24
Ex2.
f ( z) 
1
has a pole of order 2 at z0  0
z(ez 1)
Corollary: Let two functions p and q be analytic at a point z0.
If p(z0 )  0 , q(z0 )  0 , and q' (z0 )  0
then z0 is a simple pole of p(z) and
q(z)
p( z0 )
Re s p( z) 
z  z0
q( z) q' ( z0 )
Pf:
q( z)  ( z  z0 ) g ( z),
g ( z) is analytic ard non zero at z0
p( z)  p( z)/ g ( z)
z  z0
q( z)
p(z)  p( z0 )
Res
Form Theorem in sec 56,
zz
q(z) g ( z )
0
But g(z0)  q' (z0 )
tch-prob
0

p( z0 )
q' ( z0 )
25
Ex3.
f ( z)  cot z  cos z
sin z
p( z)  cos z, q(z)  sin z both entire
The singularities of f(z) occur at zeros of q, or
z  n (n  0, 1,  2, ...)
Since p(nπ)  (1)n  0, q(nπ)  0 , and q' (nπ) (-1)n  0
each singular point z  nπ of f is a simple pole,
n
with residue Bn  p(nπ)  (1)n 1
q' (nπ) (1)
try tan z
tch-prob
26
Ex4
z
z4  4
iπ
z0  2 e 4 1 i
f ( z) 
:
:
tch-prob
27
58. Conditions under which f (z)  0
Lemma : If f(z)=0 at each point z of a domain or arc
containing a point z0, then f ( z)  0 in any
neighborhood N0 of z0 throughout which f is
analytic. That is, f(z)=0 at each point z in N0.
Pf:
Under the stated condition,
Z0
arc
For some neighborhood N of z0
N
f(z)=0
N0
Otherwise from (Ex13, sec. 57)
There would be a deleted neighborhood of z0
throughout which f ( z)  0
 inconsistent with f ( z)  0 in a domain or arc
containing z0.
tch-prob
28
Since f ( z)  0 in N, an in the Taylor series for f(z) about z0
must be zero.
Thus f ( z)  0 in neighborhood N0 since that Taylor series
also represents f(z) in N0.
圖解
Z0 Z
若有一點 f ( z )  0
則全不為 0
全為 0
Z0
Z
若在 arc or domain為 0 ,則
Ex13, sec 57
Theorem. If a function f is analytic throughout a domain D
and f(z)=0 at each point z of a domain or arc
contained in D, then f ( z)  0 in D.
P
Z0 Z1 Z 2 Z3
tch-prob
Zn
29
Corollary: A function that is analytic in a domain D is uniquely
determined over D by its values over a domain, or
along an arc, contained in D.
D
f ( z), g ( z) analytic in D
f ( z)  g ( z) in some domain or arc contained in D
h( z)  f ( z) - g ( z)  0 in a domain or acr
h( z)  0 in D
f ( z)  g ( z) in D
domain
arc
Example:
Since sin2 x cos2 x 1
along real x-axis
(an arc)
f ( z)  sin 2 z  cos2 z 1
is zero along the real axis
 f ( z)  0 throughout the complex plane
 sin 2 z  cos2 z 1 for all z
tch-prob
30
59. Behavior of f near Removable and Essential
Singular Points
Observation :
A function f is always analytic and bounded in some
deleted neighborhood 0  z  z0  ε of a removable
singularity z0.
For f is analytic in a disk z- z0  R2 when f ( z0 ) is
properly defined at such a point; and f is then continuous
in any closed disk z- z0  ε when ε  R2.
Consequently, f is bounded in that disk.
(From sec. 14,) f must be bounded in 0  z- z0  ε.
tch-prob
31
Thm 1: Suppose that a function f is analytic and bounded in some
deleted neighborhood
of a point z0.
0  z- z0  
If f is not analytic at z0, then it has a removable singularity
there.
Pf: Assume f is not analytic at z0.
The point z0 is an isolated singularity of f and f(z) is
represented by a Laurent series


bn
f ( z)   an ( z- z0 )n  
n
n0
n1 ( z- z0 )
throughout 0  z- z0  ε
If C denotes a positively oriented circle
where
z- z0  
 
tch-prob
32
bn  1 c f ( z) dz
(n 1, 2, ...)
2π i ( z - z )n1
0
since f ( z)  M
0  z - z0  
bn  1 . M 2πρ  Mρn
2π ρ-n1
since  can be chosen arbitrarily small,  bn  0
 z0 is a remorable singularity of f .
C Z0 

Thm2. Suppose that z0 is an essential singularity of a function f,
and let w0 be any complex number. Then, for any positive
number  , the inequality
(a function assumes values arbitrarily
f ( z)  w0  
(3) close to any given number)
is satisfied at some point z in each deleted neighborhood
0  z  z0  of z0
tch-prob
33
Pf: Since z0 is an isolated singularity of f. There is a 0  z  z0 
throughout which f is analytic.
Suppose (3) is not satisfied for any z there. Thus
f ( z)  w0   when 0  z - z0  
g ( z) 
1
f (z)  w0 is bounded and analytic in 0  z  z0 
According to Thm 1, z0 is a removable singularity of g. We
let g be defined at z0 so that it is analytic there,
If g ( z0 )  0,
f (z)  1  w0
g ( z)
0  z  z0  δ
becomes analytic at z0 if it is defined there as
f (z0 )  1  w0
g(z0 )
But this means that z0 is a removable singularity of f, not
an essential one, and we have a contradiction.
tch-prob
34
If g ( z0 )  0, g ( z) must have a zero of some finite order m (sec. 57)
at z0 because g is not identically equal to zero in z  z0  δ.
(
f ( z)  w0   when 0  z - z0   )
In view of f ( z)  1  w0 , f has a pole of order m at z0.
g ( z)
So, again, we have a contradiction. Theorem 2 is proved.
tch-prob
35